limit of the sequence $x_{n}:= \sqrt[n]{n \sqrt[n]{n \sqrt[n]{n\ldots}}}$ I was thinking what happens with the sequence $\{x_n\}_{n\in \Bbb N}$ where:
$$x_{n}:= \sqrt[n]{n \sqrt[n]{n \sqrt[n]{n\ldots}}}$$
When you look some terms, for example $x_{1}=1$, $x_{2}=\sqrt[]{2 \sqrt[]{2 \sqrt[]{2 ...}}}$, $x_{3}=\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}$, these terms and the others will be continued fractions, where each one converges.
I'm asking what happens with $\lim\limits_{n \to \infty}\sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}$ ?. I have an idea and it is $\lim\limits_{n \to \infty}\sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}=1$. My reasoning is in the fact:
$$\sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}= \displaystyle {n^{\frac{1}{n}}} n^{\frac{1}{n^2}} n^{\frac{1}{n^3}}...$$
And you know that:
$${\displaystyle \frac{1}{n}> \frac{1}{n^k} \textrm{ for } n,k \in \Bbb N}$$
Then:
$$n^{\frac{1}{n}}> n^{\frac{1}{n^k}} \geq 1$$
Like $\lim\limits_{n \to \infty}n^{\frac{1}{n}}=1$  and $\lim\limits_{n \to \infty}1=1$, by the Squeeze Theorem $\lim\limits_{n \to \infty}\sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}=1$. Is this reasoning correct? What do you think about $x_{n}$? Do you think there is another way to prove it? I receive suggestions or comments. Thank you.
 A: Yes, the limit is $1$:
$$x_n=n^{\sum_{k=1}^{\infty}\frac{1}{n^k}}= n^{\frac{1}{n-1}}=e^{\frac{\ln(n)}{n-1}}\to 1.$$
A: An Inductive idea is:
$$x_{2}=\sqrt[]{2 \sqrt[]{2 \sqrt[]{2 ...}}}\to 2\\
x_{3}=\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}\to \sqrt[2]3\\
x_{4}=\sqrt[4]{4 \sqrt[4]{4 \sqrt[4]{4 ...}}}\to \sqrt[3]4\\
x_{5}=\sqrt[5]{5 \sqrt[5]{5 \sqrt[5]{5 ...}}}\to \sqrt[4]5\\\vdots\\
x_{n}= \sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}\to \sqrt[n-1]n$$and it tends to $$\sqrt[n-1]n=n^{\frac{1}{n-1}}\to 1$$
Implicit : idea to solve for example$$\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}=a\to \text{to the power of 3}\\a^3=3\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}\\a^3=3\underbrace{\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}}_{a}\\a^3=3a\underbrace{\to}_{a\neq 0}a^2=3\to a=\sqrt 3$$
A: Other way is: if we guarantee  that the limit exists then:
$$\lim_{n\to \infty}\sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}=L$$
$$\lim_{n\to \infty}\sqrt[n]{n}L^{1/n}=L$$
concludes that $L= 1$
A: We want to show that
$n^{1/(n-1)}
\to 1
$.
Since
$(1+1/\sqrt{n})^n
\ge 1+\sqrt{n}
\gt \sqrt{n}$
by Bernoulli's inequality,
$n^{1/n}
\lt (1+1/\sqrt{n})^2
\lt 1+3/\sqrt{n}
$
Since
$(1+x)^n \ge 1+nx
$,
$(1+nx)^{1/n}
\le 1+x
$
or
$(1+x)^{1/n}
\lt 1+x/n
$.
Therefore
$\begin{array}\\
n^{1/(n-1)}
&=n^{1/n-1/n+1/(n-1)}\\
&=n^{1/n}n^{-1/n+1/(n-1)}\\
&=n^{1/n}n^{1/(n(n-1))}\\
&\lt (1+3/\sqrt{n})(1+(n-1))^{1/(n(n-1))}\\
&\le (1+3/\sqrt{n})(1+(n-1)/(n(n-1))\\
&= (1+3/\sqrt{n})(1+1/n)
&\to 1\\
\end{array}
$
A: We have
$$1\le x_n = n^{1/n+1/n^2 +1/n^3 +\cdots + 1/n^n} \le n^{1/n+(n-1)/n^2}\le n^{2/n} = (n^{1/n})^2 \to 1^2 = 1.$$
By the squeeze theorem the limit is $1.$
